Answer :

Given: - Equations are: –


5x – 7y + z = 11


6x – 8y – z = 15


3x + 2y – 6z = 7


Tip: - Theorem – Cramer’s Rule


Let there be a system of n simultaneous linear equations and with n unknown given by







and let Dj be the determinant obtained from D after replacing the jth column by



Then,


provided that D ≠ 0


Now, here we have


5x – 7y + z = 11


6x – 8y – z = 15


3x + 2y – 6z = 7


So by comparing with theorem, lets find D , D1 and D2



Solving determinant, expanding along 1st Row


D = 5[( – 8)( – 6) – ( – 1)(2)] – 7[( – 6)(6) – 3( – 1)] + 1[2(6) – 3( – 8)]


D = 5[48 + 2] – 7[ – 36 + 3] + 1[12 + 24]


D = 250 – 231 + 36


D = 55


Again, Solve D1 formed by replacing 1st column by B matrices


Here




Solving determinant, expanding along 1st Row


D1 = 11[( – 8)( – 6) – (2)( – 1)] – ( – 7)[(15)( – 6) – ( – 1)(7)] + 1[(15)2 – (7)( – 8)]


D1 = 11[48 + 2] + 7[ – 90 + 7] + 1[30 + 56]


D1 = 11[50] + 7[ – 83] + 86


D1 = 550 – 581 + 86


D1 = 55


Again, Solve D2 formed by replacing 2nd column by B matrices


Here




Solving determinant, expanding along 1st Row


D2 = 5[(15)( – 6) – (7)( – 1)] – 11[(6)( – 6) – ( – 1)(3)] + 1[(6)7 – (15)(3)]


D2 = 5[ – 90 + 7] – 11[ – 36 + 3] + 1[42 – 45]


D2 = 5[ – 83] – 11( – 33) – 3


D2 = – 415 + 363 – 3


D2 = – 55


And, Solve D3 formed by replacing 3rd column by B matrices


Here




Solving determinant, expanding along 1st Row


D3 = 5[( – 8)(7) – (15)(2)] – ( – 7)[(6)(7) – (15)(3)] + 11[(6)2 – ( – 8)(3)]


D3 = 5[ – 56 – 30] – ( – 7)[42 – 45] + 11[12 + 24]


D3 = 5[ – 86] + 7[ – 3] + 11[36]


D3 = – 430 – 21 + 396


D3 = – 55


Thus by Cramer’s Rule, we have




x = 1


again,




y = – 1


and,




z = – 1


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